Using Theorem 1.6 (http://www.ugr.es/~acegarra/Rotman.pdf page 17), show (using the notation of exercise 3.2 (page 40)) that if $f$ and $g$ are paths with $g$ constant, then $f'\simeq g'$ rel$\{1\}$ if an only if there exist a free homotopy $f'\simeq g'.$
Attempt to the proof
$\implies$ direction
Follows immediately.
$\Longleftarrow$ direction
Suppose we have an homotopy $H:I\times I\to X$ such that $$H(t,0)=f'(t)$$ $$H(t,1)=g'(t)$$
We only need to find $H$ such that $H(1,t)=f'(1)=g'(1).$
Applying exercise 3.2 i) to $f$ and $g$, I have $f'\simeq g'.$ Since $g'$ is now constant, then $f'$ is nullhomotopic.
Now I am not sure if I can use iii) from Theorem 1.6 because I have $f':S^1\to X$ and I need $f':S^n\to X$. I am wrong in the previous line, in fact I can use the Theorem 1.6 with $n=1.$
I think now I can conclude that there exist an homotopy $H:f'\simeq g'$ such that $H(cte,t)=f'(cte),$ for all $t\in I.$
That is $H(1,t)=f'(1)$ for all $t\in I$.
I can have $H(1,t)=f'(1)=g'(1)$ only if $H(t=1,1).$ How can I correct this?
Am I on the right track?
If someone could help me, thank you.
You confuse $f,g : I \to X$ which are closed paths and $f',g' : S^1 \to X$. Thus your homotopy $H : I \times I \to X$ cannot be a homotopy from $f'$ to $g'$. So take a (free) homotopy $H : S^1 \times I \to X$ such that $H(z,0) = f'(z), H(z,1) = g'(z) = x_0$.
In Example 1.7 Rotman shows that if $f : S^n \to X$ is map and $F : S^n \times I \to X$ a homotopy such that $F(z,0) = f(z), F(z,1) = \xi$ for all $x$, then $f$ has a continuous extension $F' : D^{n+1} \to X$.
So let $H': D^2 \to X$ be a continuous extension of $f'$. Define $$R : S^1 \times I \to D^2, R(z,t) = t \cdot 1 + (1-t)\cdot z$$ where we regard $D^2$ as a subset of the complex plane. The map $R$ is continuous. Define $$H'' = H' \circ R : S^1 \times I \to X .$$ Then $H''(z,0) = H'(z) = f'(z)$, $H''(z,1) = H'(1) = f'(1) = x_0 = g'(z)$ and $H''(1,t) = H'(1) = x_0$. Thus $H''$ is the desired basepoint-preserving homotopy.